# The Naked Scientists Forum

### Author Topic: A numerical problem..  (Read 3280 times)

#### hamza

• Full Member
• Posts: 88
##### A numerical problem..
« on: 20/09/2007 15:35:27 »
A bullet of mass 10 gram strikes a fix target and penetrates 8 cm into it. if the average resistance offered by the target to the bullet is 100 Newton. What's  the velocity with which the bullet hits the target.
« Last Edit: 20/09/2007 15:45:11 by hamza »

#### eric l

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• Posts: 514
##### A numerical problem..
« Reply #1 on: 20/09/2007 16:38:33 »
Are we making your homework ?

We have a mass and a force (actually a resistance, but that amounts to the same thing).  We can calculate the accelaration (because it is a negative force, the acceleration will be negative, too).
Force = mass * acceleration or
acceleration = force / mass = 100 N / 0.01 kg = 10000 m/sec2

Now we cheat a bit, and make the movement in reverse.  We start with speed zero, and accelerate till we are out of the target.
When starting from zero, the distance covered is
(1/2)*acceleration*time2
distance is 0.08 m; acceleration is 10000 m/sec2
time2 = (0.16m)/(10000m/sec2) = (16/1000000)sec2
time = 4/1000 sec
speed = (0.004 sec)*(10000 m:sec2) = 40 m/sec

This must be equal to the speed at impact.

#### hamza

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• Posts: 88
##### A numerical problem..
« Reply #2 on: 20/09/2007 17:37:54 »
Are we making your homework ?
Thanks alot Eric.. and NO, this is definitely not my homework. I had this one in a numericals book and was confused so i thought about asking.. Thanks alot once again

#### eric l

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• Posts: 514
##### A numerical problem..
« Reply #3 on: 20/09/2007 18:19:00 »
Have you seen the formulas for accelerated movement before or were they new fopr you, too ?

#### lightarrow

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##### A numerical problem..
« Reply #4 on: 20/09/2007 19:53:43 »
A bullet of mass 10 gram strikes a fix target and penetrates 8 cm into it. if the average resistance offered by the target to the bullet is 100 Newton. What's  the velocity with which the bullet hits the target.

Work done on the bullet: F•S = 100 N•0.08 m = 8 Nm
Kinetic energy of the bullet = work done on it:
(1/2)Mv2 = 8 Nm --> v2 = 16/M = 16Nm/0.01Kg = 1600 m2/s2 --> v = 40 m/s.

#### eric l

• Hero Member
• Posts: 514
##### A numerical problem..
« Reply #5 on: 20/09/2007 19:57:58 »
Lightarrow's approach to the problem is the correct one, but I thought that hanza would be more familiar with the notions about accelerated movement.
And of course, we arrive at the same result.

#### hamza

• Full Member
• Posts: 88
##### A numerical problem..
« Reply #6 on: 21/09/2007 17:03:23 »
Have you seen the formulas for accelerated movement before or were they new fopr you, too ?
Yes, I am familiar with the formulas you have used, but i was confused about their proper application. Anyways, Eric! your solution helped alot.
And lightarrow, thanks to you too as your solution is more simpler.. (no offence to eric here..) Both of the solutions have helped me and provided me with a broader perspective for the solution.. Thanks again..

#### The Naked Scientists Forum

##### A numerical problem..
« Reply #6 on: 21/09/2007 17:03:23 »